NCERT New Syllabus
Mathematics
Class: 10
Exercise 1.1
Topic: Real
Numbers
Understanding Real Numbers: Key to Mathematical Mastery.
Embark on a journey to explore a crucial subject in mathematics: real numbers. This section from the NCERT class 10 curriculum provides the basis for understanding more complex mathematical ideas by focusing on properties, theorems, and operations involving real numbers. This article will delve into exercise 1.1, thoroughly breaking down each problem and providing solutions. Whether you are preparing for an examination or building a strong foundation, this guide will assist you in navigating each answer effortlessly. Let's delve into the significance of real numbers together!
1) Theorem 1.1 (Fundamental Theorem of Arithmetic):
Every composite number may be written as a unique product of prime numbers, regardless of how the prime elements are organized. Each composite number has a unique prime factorization except the prime number sequence.
2) For any two positive integers a and b, the relationship between their HCF and
LCM is given by HCF (a, b) × LCM (a, b) = a × b.
3) This formula allows us to calculate the LCM of two positive integers easily
we know their HCF.
EXERCISE 1.1
Q1. Express each number as a product of its prime factors:
(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
Explanation:
Every composite number can be represented as a product of prime numbers.
Solution:
(i) 140
1) Prime factor of 140:
Note: Apply the divisibility test of 2.
140 = 2 x 70 Note: Apply the divisibility test of 2.
140 = 2 x 2 x 35 Note: Apply the divisibility test of 5.
140 = 2 x 2 x 5 x 7
140 = 22 x 5 x 7
2) The prime factorization of 140 is 22 x 5 x 7.
(ii) 156
1) Prime factor of 156:
Note: Apply divisibility test of 2.
156 = 2 x 78 Note: Apply divisibility test of 2.
156 = 2 x 2 x 39 Note: Apply divisibility test of 3.
156 = 2 x 2 x 3 x 13
156 = 22 x 3 x 13
2) The prime factorization of 156 is 22 x 3 x 13.
(iii) 3825
1) Prime factor of 3825: Note: Apply the divisibility test of 3.
3825 = 3 x 1275 Note: Apply divisibility test of 3.
3825 = 3 x 3 x 425 Note: Apply divisibility test of 5.
3825 = 3 x 3 x 5 x 85 Note: Apply divisibility test of 5.
3825 = 3 x 3 x 5 x 5 x 17
3825 = 32 x 52 x 17
2) The prime factorization of 3825 is 32 x 52 x 17.
(iv) 5005
1) Prime factor of 5005: Note: Apply the divisibility test of 5.
5005 = 5 x 1001 Note: Apply divisibility test of 7.
5005 = 5 x 7 x 143 Note: Apply divisibility test of 11.
5005 = 5 x 7 x 11 x 13
2) The prime factorization 5005 is 5 x 7 x 11 x 13.
(v) 7429
Note:
a) 2 is not a factor of 7429 as it is not an even number. So 4, 8, 16, and so on
are also not the factors of 7429.
b) 3 is not a factor of 7429 as the sum of the digits is 22, which is not divisible
by 3. So 6, 9, 12, 15, and so on are also not factors 7429.
c) 5 is not a factor of 7429 as the unit placed digit is not 0 or 5. So 5, 10, 15,
and so on are also not factors 7429.
d) 7 is not a factor of 7429. So 14, 21, and so on are also not factors 7429.
1) Prime factor of 7429: Note: Try directly dividing by 17.
7429 = 17 x 437 Note: Try directly dividing by 19.
7429 = 17 x 19 x 23
2) The prime factorization of 7429 is 17 x 19 x 23.
Q2. Find the LCM and HCF of the following pairs of integers and verify that
LCM × HCF = product of the two numbers.
(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54
Explanation:
1) To determine the LCM and HCF of given numbers, we first need to identify their
prime factors.
2) The HCF (Highest Common Factor) is found by multiplying the smallest powers
of the common prime factors between the numbers.
3) The LCM (Least Common Multiple) is calculated by multiplying the greatest
powers of each prime factor present in the numbers.
4) The relationship between HCF and LCM is given by the formula:
HCF (p, q) × LCM (p, q) = p × q, where p and q are any two numbers that are positive integers.
Solution:
(i) 26 and 91
1) First, find the prime factors of 26:
26 = 2 x 13
2) Next, find the prime factors of 91:
91 = 7 x 13
3) The common factor between 26 and 91 is 13.
4) Therefore, the HCF of 26 and 91 is 13.
5) To get the LCM, multiply the individual prime factors:
LCM = 2 x 7 x 13
LCM = 14 x 13
LCM = 182.
6) Thus, the LCM of 26 and 91 is 182.
7) Therefore, the LCM of 26 and 91 is 182, and the HCF of 26 and 91 is 13.
8) We know 26 x 91 = 2366 -------- (equation 1).
9) Also, HCF x LCM = 13 x 182 = 2366 -------- (equation 2).
10) From Equations 1 and 2, we confirm that:
HCF(26,91) x LCM(26,91) = 26 x 91.
(ii) 510 and 92
1) First, find the prime factors of 510:
510 = 2 x 255
510 = 2 x 3 x 85
510 = 2 x 3 x 5 x 17
2) Next, find the prime factors of 92:
92 = 2 x 4692 = 2 x 2 x 23
3) The common factor between 510 and 92 is 2.
4) Therefore, the HCF of 510 and 92 is 2.
5) To get the LCM, multiply the individual prime factors:
LCM = 2 x 2 x 3 x 5 x 17 x 23.
LCM = 4 x 15 x 17 x 23.
LCM = 23460.
6) Thus, the LCM of 510 and 92 is 23460.
7) Therefore, the LCM of 510 and 92 is 23460, and the HCF of 510 and 92 is 2.
8) We know 510 x 92 = 46920 -------- (equation 1)
9) Also, HCF x LCM = 2 x 23460 = 46920 -------- (equation 2).
10) From Equations 1 and 2, we confirm that:
HCF(510,92) x LCM(510,92) = 510 x 92.
(iii) 336 and 54
1) First, find the prime factors of 336.
336 = 2 x 168336 = 2 x 2 x 84336 = 2 x 2 x 2 x 42336 = 2 x 2 x 2 x 2 x 21336 = 2 x 2 x 2 x 2 x 3 x 7
2) Next, find the prime factors of 54.
54 = 2 x 2754 = 2 x 3 x 954 = 2 x 3 x 3 x 3
3) The common factor between 336 and 54 is 2 x 3 = 6.
4) Therefore, the HCF of 336 and 54 is 6.
5) To get the LCM, multiply the individual prime factors:
LCM = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 7
LCM = 16 x 27 x 7
LCM = 3024.
6) Thus, the LCM of 336 and 54 is 3024
7) Therefore, the LCM of 336 and 54 is 3024, and the HCF of 336 and 54 is 6.
8) We know 336 x 54 = 18144 -------- (equation 1).
9) Also, HCF x LCM = 6 x 3024 = 18144 -------- (equation 2).
10) From Equations 1 and 2, we confirm that
HCF(336, 54) x LCM(336, 54) = 336 x 54.
Q3. Find the LCM and HCF of the following integers by applying the prime factorization
factorization method.
(i) 12, 15, and 21 (ii) 17, 23, and 29 (iii) 8, 9, and 25
Solution:
(i) 12, 15, and 21
1) Now we will find the prime factors of 12.
12 = 2 x 612 = 2 x 2 x 3
2) Now we will find the prime factors of 15.
15 = 3 x 5
3) Now we will find the prime factors of 21.
21 = 3 x 7
4) Here, the common factor is 3.
5) The HCF of 12, 15, and 21 is 3.
6) Similarly, LCM = 2 x 2 x 3 x 5 x 7 = 420.
7) Thus, LCM is 420, and HCF is 3.
(ii) 17, 23, and 29
1) Now we will find the prime factors of 17.
17 = 1 x 17
2) Now we will find the prime factors of 23.
23 = 1 x 23
3) Now we will find the prime factors of 29.
29 = 1 x 29
4) The common factor is 1.
5) The HCF of 17, 23, and 29 is 1.
6) Similarly, LCM = 1 x 17 x 23 x 29 = 11339.
7) Thus, LCM is 11339, and HFC is 1.
(iii) 8, 9, and 25
1) Now we will find the prime factors of 8.
8 = 1 x 2 x 48 = 1 x 2 x 2 x 2
2) Now we will find the prime factors of 9.
9 = 1 x 3 x 3
3) Now we will find the prime factors of 25.
25 = 1 x 5 x 5
4) The common factor is 1.
5) So the HCF of 8, 9, and 25 is 1.
6) Similarly, LCM = 1 x 2 x 2 x 2 x 3 x 3 x 5 x 5 = 1800.
7) Thus, LCM is 1800, and HCF is 1.
Q4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Explanation:
1) We know that HCF (p, q) × LCM (p, q) = p × q, where p and q are any two numbers that are positive integers.
2) Out of the above 3 values, if 2 values are given, we can find the 3rd value.
Solution:
1) Here, HCF (306, 657) = 9.
2) So, the HCF of 306 and 657 = 9.
3) Products of 306 and 657 are 306 x 657
4) We know that HCF (p, q) × LCM (p, q) = p × q, so,
[LCM (306, 657)] = [306 x 657] / HCF (306, 657)
[LCM (306, 657)] = [306 x 657] / 9
[LCM (306, 657)] = [34 x 657][LCM (306, 657)] = [22338]
5) Thus, LCM (306, 657) = 22338.
Q5. Check whether 6n can end with the digit 0 for any natural number n.
Explanation:
1) If any number ends with the digit 0, that number must be divisible by 2 and 5 simultaneously.
2) To check whether 6n ends with 0, we must find whether 6n has 2 and 5 as prime factors.
Solution:
1) Factors of 6n = (2 x 3)n
2) Here 2 is the factor of 6n, but 5 is not a factor of 6n.
3) So 6n can't end with 0 for any natural number n.
Q6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Explanation:
1) Prime numbers have only 2 factors. 1 and the number itself. For example, the prime number 5 has two factors. 1 and 5 itself.
2) The positive number with factors other than 1 is a composite number.
Solution:
1) The first expression is (7 × 11 × 13) + 13. We will simplify it.
(7 × 11 × 13 + 13) = 13 ((7 x 11 x 1) + 1)
= 13 (77 + 1)= 13 (78)= 13 (2 x 39)= 13 (2 x 3 x 13)
2) The first expression has factors as (2 x 3 x 13 x 13)
3) The number obtained from (7 × 6 × 5 × 4 × 3 × 2 × 1) + 5 is a composite number.
4) The second expression is (7 × 11 × 13) + 13. We will simplify it.
(7 × 6 × 5 × 4 × 3 × 2 × 1) + 5 = 5 ((7 × 6 × 4 × 3 × 2 × 1) + 1)
= 5 (1008 + 1)= 5 (1009)
5) The second expression has factors as (5 x 1009)
6) The number obtained from (7 × 6 × 5 × 4 × 3 × 2 × 1) + 5 is a composite
number.
Q7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting point?
Explanation:
1) Sonia takes 18 minutes and Ravi takes 12 minutes to complete one round.
2) We need to find out how many minutes they will meet again at the same point.
3) Here we will find the LCM of 18 and 12 to get the time when both meet again at the starting point.
Solution:
1) To find the LCM of 18 and 12, we must first determine their prime factors.
2) Prime factorization of 18:
18 = 2 x 918 = 2 x 3 x 3
3) Prime factorization of 12:
12 = 2 x 612 = 2 x 2 x 3
4) The common prime factor is 2 x 3 = 6.
5) The LCM (Least Common Multiple) is the product of all the highest powers of the
prime factors:
LCM = 2 x 2 x 3 x 3 = 36.
6) Therefore, the LCM is 36.
7) After 36 minutes, Sonia and Ravi will meet again at the starting point.
Conclusion: Unveiling the Power of Real Numbers
In this chapter on real numbers, we've explored the foundational blocks of mathematics, laying the groundwork for a deeper understanding of algebra and beyond. From the beauty of irrational numbers to the precision of the Euclidean algorithm, real numbers are at the core of every mathematical operation. As you delve into the world of numbers, remember that these principles stretch far beyond the classroom, influencing technology, science, and everyday calculations. Keep exploring, calculating, and unlocking the endless possibilities of real numbers!
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